Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
const int NMAX = 1e5 + 11;
const int AMAX = 10;
struct segtree {
  int n;
  vector<vi> t;
  segtree() {}
  segtree(int n) : n(n) { t.resize(4 * n); }
  void build(vi a[]) { build_util(a, 1, 0, n - 1); }
  void build_util(vi a[], int v, int tl, int tr) {
    if (tl == tr) {
      sort(a[tl].begin(), a[tl].end());
      t[v] = a[tl];
      if (AMAX < (int)t[v].size()) t[v].resize(10);
    } else {
      int tm = (tl + tr) >> 1;
      build_util(a, v << 1, tl, tm);
      build_util(a, (v << 1) + 1, tm + 1, tr);
      t[v].resize(t[v << 1].size() + t[(v << 1) + 1].size());
      merge(t[v << 1].begin(), t[v << 1].end(), t[(v << 1) + 1].begin(),
            t[(v << 1) + 1].end(), t[v].begin());
      if (AMAX < (int)t[v].size()) t[v].resize(10);
    }
  }
  vi query(int u, int v) { return query_util(1, 0, n - 1, u, v); }
  vi query_util(int v, int tl, int tr, int ql, int qr) {
    if (ql > qr) return vi();
    if (tl == ql and tr == qr) return t[v];
    int tm = (tl + tr) >> 1;
    vi lans = query_util(v << 1, tl, tm, ql, min(tm, qr));
    vi rans = query_util((v << 1) + 1, tm + 1, tr, max(tm + 1, ql), qr);
    vi ans(lans.size() + rans.size());
    merge(lans.begin(), lans.end(), rans.begin(), rans.end(), ans.begin());
    return ans;
  }
} st;
vi g[NMAX];
vi population[NMAX];
int parent[NMAX], depth[NMAX];
int heavy[NMAX], pos[NMAX], head[NMAX], hldcnt;
int dfs(int u) {
  int usz, vsz, msz;
  usz = 1;
  msz = 0;
  for (auto v : g[u]) {
    if (parent[u] == v) continue;
    parent[v] = u;
    depth[v] = depth[u] + 1;
    vsz = dfs(v);
    usz += vsz;
    if (vsz > msz) {
      heavy[u] = v;
      msz = vsz;
    }
  }
  return usz;
}
void decompose(int u, int h) {
  head[u] = h;
  pos[u] = hldcnt++;
  if (heavy[u] != -1) decompose(heavy[u], h);
  for (auto v : g[u]) {
    if (parent[u] != v and heavy[u] != v) decompose(v, v);
  }
}
void query(int u, int v, int a) {
  int n;
  vi ans, cur;
  while (head[u] != head[v]) {
    if (depth[head[u]] > depth[head[v]]) swap(u, v);
    cur = st.query(pos[head[v]], pos[v]);
    v = parent[head[v]];
    n = ans.size();
    ans.resize(n + cur.size());
    copy(cur.begin(), cur.end(), ans.begin() + n);
    inplace_merge(ans.begin(), ans.begin() + n, ans.end());
  }
  if (depth[u] > depth[v]) swap(u, v);
  cur = st.query(pos[u], pos[v]);
  n = ans.size();
  ans.resize(n + cur.size());
  copy(cur.begin(), cur.end(), ans.begin() + n);
  inplace_merge(ans.begin(), ans.begin() + n, ans.end());
  int k = min(a, (int)ans.size());
  cout << k << " ";
  for (int i = 0; i < k; ++i) {
    cout << ans[i] + 1 << " ";
  }
  cout << endl;
}
void reset(void) {
  hldcnt = 0;
  memset(heavy, -1, NMAX * sizeof(*heavy));
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, q, u, v, ci, a;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; ++i) {
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  reset();
  dfs(0);
  decompose(0, 0);
  for (int i = 0; i < m; ++i) {
    cin >> ci;
    population[pos[ci - 1]].push_back(i);
  }
  st = segtree(n);
  st.build(population);
  while (q--) {
    cin >> u >> v >> a;
    query(u - 1, v - 1, a);
  }
  return 0;
}